How would you determine whether this sequence transformation has an inverse? Let $T : a \mapsto b$ be a transformation of sequence $a$ to $b$ of the form 
$$
T(a)_m = b_m = \sum_{k=1}^{\infty} a_k e^{-i 2 \pi m / k }
$$
Question.
How would you go about determining if this transformation is invertible?  If it is invertible, then can it be written in the same form (linear combination).
Note this is not the Fourier transform of a sequence which results in a continuous function on $\mathbb{C}$, and it is not a restriction of that map either, notice that we have $m / k$ not the opposite, which would then indeed be the same as the official Fourier transform.
I'm new to functional analysis so have know idea where to begin, but don't be afraid to use high-level terminology especial high-level in topology language since I know some of that.  
Thanks.
 A: One of the most important warning in Functional Analysis is specifying the spaces we are dealing with. Most probably, in this case $T$ is a map on $\ell^1( \mathbb N)$, i.e. $T: \ell^1( \mathbb N) \to \ell^1( \mathbb N )$. We shall assume this hypotesis. Now, we see that $T$ is an isometry, since $\| T a \|_1 = \| a \|_1$ for all $a \in \ell^1( \mathbb N )$. ($T$ acts as a multiplication for a phase factor). Moreover, $T$ is a surjection. Pick $a \in \ell^1( \mathbb N)$. Then $\sum_{n = 0}^{\infty} \lvert a_n \rvert < \infty$ and if we multiply each entry of $a$ times a complex number of absolute value smaller or equal to 1, as $e^{-i 2 \pi m / k}$, we are sure the new vector is still in $\ell^1(\mathbb N)$. Thanks to this observation, it is clear that every vector in $\ell^1(\mathbb N)$ can be achieved applying $T$ on some suitable vector of $\ell^1( \mathbb N)$. Then $T$ is invertible. Furthermore, $T^{-1}$ is bounded thanks to Banach' isomorphism theorem. (often included as a part of the statement of the open mapping theorem.) Hence $T$ is more than invertible and more than an isometry, it is an isometric isomorphism onto $\ell^1(\mathbb N)$.
Remark. I may be wrong in assumption about domain and codomain. In $c_0(\mathbb N)$ argument would hold the same way, but the result could be true even relaxing the request of absolute convergence of the series. (in analogy to the notorius case of the series $\sum_n (-1)^n/n$, which is not absolutely convergent but convergent by Leibniz' criterion.) Anyway, one always has to specify the space in which operates and the norm on it.
